Reputation: 197
I have this database :
Table User : username, password
Table Product : productID, productName
In my admin.php, if I click button with id btn, I want to display insert.php in a div with id show
admin.php :
<?php
session_start();
include("connection.php");
?>
<html>
<head>
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#btnInsert").click(function(){
$("#show").load("insert.php");
});
$("#btnEdit").click(function(){
$("#show").load("edit.php");
});
$("#btnAdd").click(function(){
$("#show").load("insertbrg.php");
});
});
</script>
</head>
<body>
<div id = "menu">
<button id = "btn">Insert</button><BR>
</div>
<div id = "show">
</div>
</body>
</html>
And in insert.php :
<?php
session_start();
include("connection.php");
?>
Product ID : <input type = "text" name = "txtID"><BR>
Product Name : <input type = "text" name = "txtName"><BR>
<input type = "button" name = "btnAdd" id = "btnAdd" value = "Add Item to DB">
if I click btnAdd, the jquery in admin.php will load insertbrg.php to add Item to database
insertbrg.php :
<?php
if(isset($_REQUEST['btnAdd'])){
$newID= $_REQUEST['txtID'];
$newName= $_REQUEST['txtName'];
$query = "Select * from Product";
$result = mysql_query($query,$conn);
$idExist = false;
if($result){
while ($row= mysql_fetch_array($result))
{
if ($row["productID"] == $newID){
$idExist = true;
}
}
}
if ($idExist){
echo "ID Exist ";
}else{
$query2="insert into product values('$newID','$newName')";
$result = mysql_query($query2,$conn);
header("location:insert.php");
}
}
?>
I can't insert the item to my db if I clicked the btnAdd button
And also I get this warning :
event.returnValue is deprecated. Please use the standard event.preventDefault() instead.
everytime I click the btn button in admin.php
Anyone know what's wrong? I'm a newbie in JQuery AJAX. Please Help...
Upvotes: 0
Views: 370
Reputation: 171
update your jQuery file and also make sure you pass the data you want to insert correctly so php can grab them and insert into the database.
you can send the data by get or post and its better to use the ajax function in jQuery.
you can find details about that here: http://api.jquery.com/jQuery.ajax/
Upvotes: 1
Reputation: 778
it's because nothing was sent in
$("#show").load("insertbrg.php");
it must accept parameters such as
$("#show").load("insertbrg.php?txtid="+txtid);
Upvotes: 1